Question

Find the limit by evaluating the derivative of a suitable function at an appropriate point? lim t approaches 0 (1-(1+t)^2/t(1+t)^2) Thank you very much

Do you mean `lim_(t to 0)(1-(1+t)^2/(t(1+t)^2))`

Answer 1

`color(blue)[lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))=lim_(t to 0)[-2(1+t)]/[t*2(1+t)+(1+t)^2]=-2/1=-2]`

Explanation:

I assume it is `lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))`

`lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))=0/0`

since the direct compensation product equal `0/0`
we will use L'hospital Rule.
L'hospital Rule `color(red)[lim_(trarra)(f'(x))/(g'(x))]`

now lets applied L'hospital Rule.

`lim_(t to 0)((1-(1+t)^2)/(t(1+t)^2))=lim_(t to 0)[-2(1+t)]/[t*2(1+t)+(1+t)^2]=-2/1=-2`

Answer 2

`lim_(t to 0) (1-(1+t)^2)/(t(1+t)^2) = -2`

Explanation:

If we seek the value of the corrected limit:

`L = lim_(t to 0) (1-(1+t)^2)/(t(1+t)^2)`

Then by expanding the numerator, we have:

`L = lim_(t to 0) (1-(1+2t+t^2))/(t(1+t)^2)`

`\ \ = lim_(t to 0) (-2t-t^2)/(t(1+t)^2)`

`\ \ = lim_(t to 0) (-t(2+t))/(t(1+t)^2)`

`\ \ = lim_(t to 0) (-(2+t))/((1+t)^2)`

And we can evaluate this limit by direct substitution:

`L = (-(2+0))/((1+0)^2)`
`\ \ = -2`